Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Scanner;

public class C419C {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String[] dir = sc.nextLine().split(" ");
		int r = Integer.parseInt(dir[0]);
		int c = Integer.parseInt(dir[1]);
		
		int[][] n = new int[r][c];
		
		for(int x=0; x<r; x++){
			String[] t = sc.nextLine().split(" ");
			for(int y=0; y<c; y++){
				n[x][y] = Integer.parseInt(t[y]);
			}
		}
		
		boolean im = false;
		
		for(int x=0; x<r; x++){
			int d = n[x][0] - n[0][0];
			for(int y=0; y<c; y++){
				if(n[x][y] - n[0][y] != d){
					im = true;
				}
			}
		}
		
		
		ArrayList<Integer> rowSteps = new ArrayList<Integer>();
		ArrayList<Integer> colSteps = new ArrayList<Integer>();
		
		if(im){
			System.out.print(-1);
		}
		else{
			
			if(r > c){
				for(int cc=0; cc<c; cc++){
					int min = 999;
					for(int y=0; y<r; y++){
						min = Math.min(min, n[y][cc]);
					}
					
					for(int y=0; y<r; y++){
						n[y][cc] -= min;
					}
					
					for(int x=0; x<min; x++){
						colSteps.add(cc+1);
					}
				}
				
				int[] firstCol = new int[r];
				for(int x=0; x<r; x++){
					firstCol[x] = n[x][0];
				}
				for(int x=0; x<r; x++){
					for(int z=0; z<firstCol[x]; z++){
						rowSteps.add(x+1);
					}
				}
				
				System.out.println(rowSteps.size() + colSteps.size());
				for(int x=0; x<rowSteps.size(); x++){
					System.out.println("row "+rowSteps.get(x));
				}
				for(int x=0; x<colSteps.size(); x++){
					System.out.println("col "+colSteps.get(x));
				}
			}
			else{
				for(int rr=0; rr<r; rr++){
					int min = 999;
					for(int y=0; y<c; y++){
						min = Math.min(min, n[rr][y]);
					}
					
					for(int y=0; y<c; y++){
						n[rr][y] -= min;
					}
					
					for(int x=0; x<min; x++){
						rowSteps.add(rr+1);
					}
				}
				
				int[] firstRow = n[0];
				for(int x=0; x<c; x++){
					for(int z=0; z<firstRow[x]; z++){
						colSteps.add(x+1);
					}
				}
				
				System.out.println(rowSteps.size() + colSteps.size());
				for(int x=0; x<rowSteps.size(); x++){
					System.out.println("row "+rowSteps.get(x));
				}
				for(int x=0; x<colSteps.size(); x++){
					System.out.println("col "+colSteps.get(x));
				}
			}
			
			
		}
	}
}


/*
int n = Integer.parseInt(sc.nextLine());
String[] t = sc.nextLine().split(" ");
int[] list = new int[n];
for(int x=0; x<n; x++){
	list[x] = Integer.parseInt(t[x]);
}
		
String[] dir = sc.nextLine().split(" ");
int a = Integer.parseInt(dir[0]);
int b = Integer.parseInt(dir[1]);
int c = Integer.parseInt(dir[2]);
int d = Integer.parseInt(dir[3]);
int e = Integer.parseInt(dir[4]);

int n = Integer.parseInt(sc.nextLine());
*/